When the oxide of generic metal m is heated at 25.0 °c. only a negligible amount of m is produced.

As soon as the oxide of general material M is heated at 25C, just a
minimal number of M is created.
MO2(s) <---> M(s)+O2(g)
delta G = 288.5 kJ/mol 1.) As soon as the response is paired into the transformation of graphite
to carbon-dioxide, it becomes natural.
What’s the chemical equation with this paired procedure? Show that
the response is within balance, consist of actual says, and
represent graphite as C(S) _________________________________
2.) what’s the thermodynamic balance continual the paired
response?
K = ????

(1) (A) MO2(s) <=> M(s) + O2(g), Delta G1 = 288.5
kJ/mol
(B) C(s) + O2(g) <=> CO2(g), Delta G2 = -394.4 kJ/mol

Include equation (A) + equation (B):
MO2(s) + C(s) + O2(g) <=> M(s) + O2(g) + CO2(g)

Cancel typical terms for paired response:
MO2(s) + C(s)
<=> M(s) + CO2(g)

Delta Get = Delta G1 + Delta G2
= 288.5 + (-394.4)
= -105.9 kJ/mol = -105900 J/mol

(2) heat T = 25 deg C = 298.15 K
Molar fuel continual roentgen = 8.314 J/mol.K

Equilibrium continual K = exp(-Delta Go/RT)
= exp(105900/(8.314 x 298.15))
= 3.58 x
10^(18)

(Note: if you are using T = 298 K, after that K = 3.66 x
10^18)
O2M(s)+ C(s) —-> M(s) + CO2(g)
I’d include 288.9 to dGof for CO2 of -394.4 for -105.5 kJ/mol
the response you reveal. After That
(105,500/8.314*298) = lnK and I also have and endless choice for K. Somethng
like 3.1E18

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